We are given a preference game over player set $[n]$, including preference relation $\geqprefer_i$ for all $i \in 1 \ldots n$. We will convert this into a fractional stable paths problem. Create a node $v_i$ for each $i$. Also create a universal destination node $d$. For all $i$, define $P_{ii} = $ the path $(v_i, d)$. For all $i, j$, define $P_{ij} = $ the path $(v_i, v_j, d)$. Let $\pi(v_i)$ (the set of preferred paths for $v_i$) = $\{P_{ij}: j \geqprefer_i i\}$. If $k \geqprefer_i j$, then $P_{ik} \geqprefer_i P_{ij}$.

Let $w_i(j)$ refer to the amount of weight placed by node $v_i$ on path $P_{ij}$ in a fractional SPP solution, and let $w_i(i)$ be the amount of weight placed by $i$ on path $P_{ii}$. Now we will show that $w$ is a fractional stable paths solution if and only if $w$ defines an equilibrium of the preference game.

\BfPara{$w$ is a fractional stable paths solution $\Rightarrow$ $w$ is an equilibrium of the preference game}
By the unity condition, for each $i$, $\sum_{j: P_{ij} \in \pi(v_i)} w_i(j) \leq 1 \Rightarrow \sum_j w_i(j) \leq 1$. $P_{ii}$ starts at $v_i$, and there is no proper final suffix of $P_{ii}$, so condition (S1) must apply for $P_{ii}$. Therefore, $\sum_{j: P_{ij} \in \pi(v_i)} w_i(j) = \sum_j w_i(j) = 1$, as required for the preference game. By the tree condition, for any $i, j$, $\sum_{P \in \pi(v, P_{jj}}$ weight on $P \leq w_j(j) \Rightarrow w_i(j) \leq w_j(j)$. So $w$ is a feasible weight assignment for the preference game.

Now suppose for contradiction that $w$ is not lexicographically maximal (with respect to $w_{-i}$) for player $i$ in the preference game. Then, there is some feasible weight assignment $w'$ and some $j$ such that $\sum_{k \geqprefer_i j} w_i(k) < \sum_{k \geqprefer_i j} w'_i(k)$.  Take the lexicographically maximal such $w'$ and the highest preference such $j$ (from $i$'s preference list).  By this choice of $w'$ and $j$, $\sum_{k \gtprefer_i j} w_i(k) = \sum_{k \gtprefer_i j} w'_i(k)$, so $\sum_{k \eqprefer_i j} w_i(k) < \sum_{k \eqprefer_i j} w'_i(k)$. There must be some $j'$ with $j' \eqprefer_i  j$ such that $w_i(j') < w'_i(j')$.
Consider path $P_{ij'}$. (S2) is not true by our choice of $j'$ and the fact that $w'$ was a feasible solution (so $w'_i(j') \leq w_{j'}(j')$). However, since $\sum_{k \eqprefer_i j'} w_i(k) < \sum_{k \eqprefer_i j'} w'_i(k)$, there must be some path $P_{ik}$ such that $k \ltprefer_i j'$ with $w_i(k) > 0$. So (S1) is also not true, and $w$ was not a stable solution - a contradiction.

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\BfPara{$w$ is an equilibrium of the preference game $\Rightarrow$ $w$ is a fractional stable paths solution}
We know that $\sum_j w_i(j) = 1$ for all $i$. This immediately satisfies the unity condition. Since $w$ is a feasible set of weights for the preference game, $w_i(j) \leq w_j(j)$ for all $i, j$. This means that the weight placed on $P_{ij}$ is at most the weight placed on $P_{jj}$. Since $P_{ij}$ is the only path from $v_i$ that passes through node $v_j$, the tree condition holds. Now consider any path $P_{ij}$ from node $i$. Case 1: $w_i(j) = w_j(j)$. In this case, condition (S2) is satisfied.  Case 2: $w_i(j) < w_j(j)$. Because $w$ was lexicographically maximal, any weight assignment $w'$ with $\sum_{k \geqprefer_i j} w'_i(k) \geq \sum_{k \geqprefer_i j} w_i(k)$ must be infeasible. We said that $w_i(j) < w_j(j)$, so it is only possible for all such $w'$ to be infeasible if $\sum_{k \geqprefer_i j} w_i(k) = 1$. Then $\sum_{k \ltprefer_i j} w_i(k) = 0$, so (S1) is satisfied.